Question

The coordinates of the foot of the perpendicular from the point (2, 3) on the line y = 3x + 4 are given by

(a) $\left(\frac{37}{10},-\frac{1}{10}\right)$

(b) $\left(-\frac{1}{10},\frac{37}{10}\right)$

(c) $\left(\frac{10}{37},-10\right)$

(d) $\left(\frac{2}{3},-\frac{1}{3}\right)$

Answer 1

Let us coordinate of for of perpendicular is D(a, b)

y = 3x + 4 ...(1)
Slope of BC is 3
$\therefore \mathrm{slope}\mathrm{of}AD\mathrm{is}\frac{-1}{3}$ ($\because$ BC and AD are perpendicular)
Hence equation of AD is y − 3 = $\frac{-1}{3}\left(x-2\right)$
i.e. 3y − 9 = − x + 2
i.e. x + 3y = 11 ...(2)
Coordinate of D are given by point of intersection of (1) and (2),
$$\begin{gathered} \mathrm{i}.\mathrm{e}.\begin{array}{c}3x-y+4=0\end{array} \\ 3x+9y-33=0 \\ \frac{\begin{array}{c}--+\end{array}}{-10y+37=0} \\ \mathrm{i}.\mathrm{e}.y=\frac{37}{10}\mathrm{and}x=11-3\left(\frac{37}{10}\right) \\ =\frac{110-111}{10}=\frac{-1}{10} \end{gathered}$$

Hence, the correct answer is option B.